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I have a question about the following result:

If $f ∈ \mathbb{Z}[X]$ is primitive and there is a prime $p$ not dividing the leading coefficient of $f$ such that $f$ is irreducible in $(\mathbb{Z}/p \mathbb{Z})[X]$ then $f(x)$ is irreducible in $\mathbb{Z}[X]$.

My question: Does this result still hold if $p$ is composite? I have looked at a few proofs of this statement and I’m not sure how the fact that $p$ is prime is being used. Any insight would be appreciated.

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  • $\begingroup$ The same question was asked here, and I suppose there are more. $\endgroup$ Mar 17, 2022 at 17:53
  • $\begingroup$ @DietrichBurde Obviously we require that the leading coefficient is a unit $\endgroup$
    – reuns
    Mar 17, 2022 at 19:23
  • $\begingroup$ Yes, but still the point is that then $\Bbb Z/n$ has zero divisors. $\endgroup$ Mar 17, 2022 at 19:26
  • $\begingroup$ @DietrichBurde No the point is that we need to restrict to polynomials whose leading coefficient is a unit and that it works well. $\endgroup$
    – reuns
    Mar 17, 2022 at 19:28
  • $\begingroup$ So the result does not hold as the OP wanted, but only with monic polynomials? The other linked question also had this problem. $\endgroup$ Mar 17, 2022 at 19:31

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